NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
AB is a light rigid rod, which is rotating about a vertical axis passing through end A . A spring of force constant k and natural length l is attached at A and its other end is attached to a small bead of mass m . The bead can slide without friction on the rod. At the initial moment, the bead is at rest (w.r.t the rod) and the spring is unstretched. Select incorrect options :
Options
- AV max = m ω 2 l 2 k - m ω 2
- BV max = m ω 4 l 2 k - m ω 2
- CV max = m ω 4 l 2 m ω 2 - k
- DV max = m ω 2 l 2 m ω 2 - k
Correct answer
B. V max = m ω 4 l 2 k - m ω 2
Step-by-step solution
Velocity will be maximum at the equilibrium position ⇒ k x = m ω 2 l + x ⇒    x = m ω 2 l k - m ω 2 Now, using work-energy theorem ∆ K E = Work done by all the forces 1 2 m V max 2 = ∫ 0 x m ω 2 l + x d x - 1 2 k x 2 ⇒    V max 2 = 2 m ω 2 l x + m ω 2 x 2 - k x 2 m V max 2 = m ω 2 l + m ω 2 l + x - k x x m = m ω 2 l x m ⇒    V max 2 = ω 2 l x = m ω 4 l 2 k - m ω 2 V max = m &#